AP EAPCET 25-Aug-2021 Shift 2 Paper

Given, latus rectum of hyperbola subtends 90° at its centre.
Let there be a hyperbola

x2

a2

−

y2

b2

=1 ...(i)
So, eccentricity, e=√1+

b2

a2

End points of latusrectum are
L=(ae,

b2

a

) and L′=(ae,

−b2

a

) Centre =(0,0)
∵ ∠LCL′=90°
Then, in ΔLCL′, by using Pythagoras theorem,
(LC)2+(L′C)2=(LL′)2
=(

2b2

a

)2
2(a2e2+

b4

a2

)=

4b4

a2

⇒

2b4

a2

=2a2e2
⇒ b4=a4e2
[a2(e2−1)]2=a4e2 {∵b2=a2(e2−1)}
a4(e2−1)2=a4e2
⇒ (e2−1)2=e2
⇒e2−1=±e
⇒ e2≠e−1=0
⇒ e=

±1±√1+4

2

=

±1±√5

2

∵ e can't be negative, hence e=

√5−1

2

√5+1

2